Three-Dimensional Rotations as Products of Simpler Rotations

Transcript

1 Physics 116A Winter 011 Three-Dimensional Rotations as Products of Simpler Rotations 1. The most general 3 3 rotation matrix In a class handout entitled, Three-Dimensional Proper and Improper Rotation Matrices, I derived the following general form for a 3 3 matrix representation of a proper rotation by an angle θ in a counterclockwise direction about a fixed rotation axis parallel to the unit vector ˆn = (n 1, n, n 3 ), cosθ + n 1(1 cosθ) n 1 n (1 cos θ) n 3 sin θ n 1 n 3 (1 cos θ) + n sin θ R(ˆn, θ) = n 1 n (1 cosθ) + n 3 sin θ cosθ + n (1 cosθ) n n 3 (1 cosθ) n 1 sin θ n 1 n 3 (1 cosθ) n sin θ n n 3 (1 cosθ) + n 1 sin θ cosθ + n 3 (1 cosθ) (1) which is called the angle-and-axis parameterization of a three-dimensional rotation. In deriving eq. (1), we showed in the class handout cited above that R(ˆn, θ) = PR(k, θ)p 1, () where P is given by P = n 3 n 1 n 1 + n n 3 n n 1 + n n n 1 + n n 1 n 1 + n n 1 n. (3) n 1 + n 0 n 3 The general rotation matrix R(ˆn, θ) given in eq. (1) satisfies the following two relations: [R(ˆn, θ)] 1 = R(ˆn, θ) = R( ˆn, θ), (4) R(ˆn, π θ) = R( ˆn, θ), (5) which implies that any proper three-dimensional rotation can be described by a counterclockwise rotation by θ about an arbitrary axis ˆn, where 0 θ π. The angle θ can be obtained from the trace of R(ˆn, θ), cosθ = 1 (Tr R 1) and sin θ = (1 cos θ) 1/ = 1 (3 Tr R)(1 + Tr R). (6) 1

2 In this convention for the range of the angle θ, the overall sign of ˆn is meaningful for 0 < θ < π. In this case, we showed in the class handout cited above that ( ) 1 ˆn = R 3 R 3, R 13 R 31, R 1 R 1, Tr R 1, 3. (3 Tr R)(1 + Tr R) In the case of θ = π (which corresponds to TrR = 1), eq. (5) implies that (7) R(ˆn, π) = R( ˆn, π), (8) which means that R(ˆn, π) and R( ˆn, π) represent the same rotation. In this case, one can use eq. (1) to derive ) 1 ˆn = (ǫ 1 (1 + R 1 11), ǫ (1 + R 1 ), ǫ 3 (1 + R 33), if Tr R = 1, (9) where the individual signs ǫ i = ±1 are determined up to an overall sign via ǫ i ǫ j = R ij (1 + Rii )(1 + R jj ), for fixed i j, R ii 1, R jj 1. (10) The ambiguity of the overall sign of ˆn is not significant in light of eq. (8). Finally, in the case of θ = 0 (which corresponds to Tr R = 3), R(ˆn, 0) = I is the 3 3 identity matrix, which is independent of the direction of ˆn.. R(ˆn, θ) expressed as a product of simpler rotation matrices In this section, we shall demonstrate that it is possible to express a general rotation matrix R(ˆn, θ) as a product of simpler rotations. This will provide further geometrical insights into the properties of rotations. First, it will be convenient to express the unit vector ˆn in spherical coordinates, ˆn = (sin θ n cosφ n, sin θ n sin φ n, cosθ n ), (11) where θ n is the polar angle and φ n is the azimuthal angle that describe the direction of the unit vector ˆn. Noting that cosφ n sin φ n 0 cosθ n 0 sin θ n R(k, φ n ) = sin φ n cos φ n 0, R(j, θ n ) = 0 1 0, sin θ n 0 cosθ n one can identify the matrix P given in eq. (3) as: cosθ n cosφ n sin φ n sin θ cosφ n P = R(k, φ n )R(j, θ n ) = cosθ n sin φ n cosφ n sin θ sin φ n. (1) sin θ n 0 cosθ n

3 Let us introduce the unit vector in the azimuthal direction, ˆϕ = ( sin φ n, cosφ n, 0). Inserting n 1 = sin φ n and n = cosφ n into eq. (1) then yields: cosθ n + sin φ n (1 cosθ n ) sin φ n cosφ n (1 cosθ n ) sin θ n cos φ n R( ˆϕ, θ n ) = sin φ n cosφ n (1 cosθ n ) cosθ n + cos φ n (1 cosθ n ) sin θ n sin φ n sin θ n cosφ n sin θ n sin φ n cosθ n = R(k, φ n )R(j, θ n )R(k, φ n ) = PR(k, φ n ), (13) after using eq. (1) in the final step. Eqs. (4) and (13) then imply that One can now use eqs. (4), () and (14) to obtain: P = R(ˆϕ, θ n )R(k, φ n ), (14) R(ˆn, θ) = PR(k, θ)p 1 = R( ˆϕ, θ n )R(k, φ n )R(k, θ)r(k, φ n )R( ˆϕ, θ n ). (15) Since rotations about a fixed axis commute, it follows that R(k, φ n )R(k, θ)r(k, φ n ) = R(k, φ n )R(k, φ n )R(k, θ) = R(k, θ), since R(k, φ n )R(k, φ n ) = R(k, φ n )[R(k, φ n )] 1 = I. Hence, eq. (15) yields: R(ˆn, θ) = R( ˆϕ, θ n )R(k, θ)r( ˆϕ, θ n ) (16) The geometrical interpretation of eq. (16) is clear. Consider R(ˆn, θ) v for any vector v. This is equivalent to R(ˆϕ, θ n )R(k, θ)r( ˆϕ, θ n ) v. The effect of R( ˆϕ, θ n ) is to rotate the axis of rotation ˆn to k (which lies along the z-axis). Then, R(k, θ) performs the rotation by θ about the z-axis. Finally, R(ˆϕ, θ n ) rotates k back to the original rotation axis ˆn. 1 One can derive one more interesting relation by combining the results of eqs. (1), (14) and (16): R(ˆn, θ) = R(k, φ n )R(j, θ n )R(k, φ n )R(k, θ)r(k, φ n )R(j, θ n )R(k, φ n ) = R(k, φ n )R(j, θ n )R(k, θ)r(j, θ n )R(k, φ n ). That is, a rotation by an angle θ about a fixed axis ˆn (with polar and azimuthal angles θ n and φ n ) is equivalent to a sequence of rotations about a fixed z and a fixed y-axis. In fact, one can do slightly better. One can prove that an arbitrary rotation can be written as: R(ˆn, θ) = R(k, α)r(j, β)r(k, γ), where α, β and γ are called the Euler angles. Details of the Euler angle representation of R(ˆn, θ) will be presented in Section 3. 1 Using eq. (13), one can easily verify that R(ˆϕ, θ n )ˆn = k and R(ˆϕ, θ n )k = ˆn. 3

4 3. Euler angle representation of R(ˆn, θ) An arbitrary rotation matrix can can be written as: R(ˆn, θ) = R(k, α)r(j, β)r(k, γ), (17) where α, β and γ are called the Euler angles. The ranges of the Euler angles are: 0 α, γ < π and 0 β π. We shall prove these statements by construction. That is, we shall explicitly derive the relations between the Euler angles and the angles θ, θ n and φ n that characterize the rotation R(ˆn, θ) [where θ n and φ n are the polar and azimuthal angle that define the axis of rotation ˆn as specified in eq. (11)]. These relations can be obtained by multiplying out the three matrices on the right-hand side of eq. (17) to obtain cosαcosβ cosγ sin α sin γ cosαcosβ sin γ sin α cosγ cosαsin β R(ˆn, θ) = sin α cosβ cosγ + cos α sin γ sin α cosβ sin γ + cosαcosγ sin α sin β. sin β cos γ sin β sin γ cosβ (18) One can now make use of the results of Section 1 to obtain θ and ˆn in terms of the Euler angles α, β and γ. For example, cosθ is obtained from eq. (6). Simple algebra yields: cosθ = cos (β/) cos(γ + α) sin (β/) (19) where I have used cos (β/) = 1 (1 + cosβ) and sin (β/) = 1 (1 cosβ). Thus, we have determined θ mod π, consistent with our convention that 0 θ π [cf. eq. (6) and the text preceding this equation]. One can also rewrite eq. (19) in a slightly more convenient form, cos θ = 1 + cos (β/) cos 1 (γ + α). (0) We examine separately the cases for which sin θ = 0. First, cos β = cos(γ + α) = 1 implies that θ = 0 and R(ˆn, θ) = I. In this case, the axis of rotation, ˆn, is undefined. Second, if θ = π then cosθ = 1 and ˆn is determined up to an overall sign (which is not physical). Eq. (0) then implies that cos (β/) cos 1 (γ + α) = 0, or equivalently (1 + cosβ) [1 + cos(γ + α)] = 0, which yields two possible subcases, (i) cosβ = 1 and/or (ii) cos(γ + α) = 1. In subcase (i), if cosβ = 1, then eqs. (9) and (10) yield cos(γ α) sin(γ α) 0 R(ˆn, π) = sin(γ α) cos(γ α) 0, where ˆn = ± ( sin 1 (γ α), cos 1 (γ α), 0). 4

5 In subcase (ii), if cos(γ + α) = 1, then cos γ + cosα = cos 1 (γ α) cos 1 (γ + α) = 0, sin γ sin α = sin 1 (γ α) cos 1 (γ + α) = 0, since cos 1 (γ + α) = 1 [1 + cos(γ + α)] = 0. Thus, eqs. (9) and (10) yield cosβ sin α sin (β/) sin(α) sin (β/) cosαsin β R(ˆn, π) = sin(α) sin (β/) 1 + sin α sin (β/) sin α sin β, cosαsin β sin α sin β cosβ where ˆn = ± ( sin(β/) cosα, sin(β/) sinα, cos(β/) ). Finally, we consider the generic case where sin θ 0. Using eqs. (7) and (18), R 3 R 3 = sin β sin 1(γ α) cos 1 (γ + α), R 13 R 31 = sin β cos 1(γ α) cos 1 (γ + α), R 1 R 1 = cos (β/) sin(γ + α). In normalizing the unit vector ˆn, it is convenient to write sin β = sin(β/) cos(β/) and sin(γ + α) = sin 1(γ + α) cos 1 (γ + α). Then, we compute: [ (R3 R 3 ) + (R 13 R 31 ) + (R 1 R 1 ) ] 1/ = 4 cos 1 (γ + α) cos(β/) sin (β/) + cos (β/) sin 1 (γ + α). (1) Hence, ˆn = ǫ sin (β/) + cos (β/) sin 1 (γ + α) ) (sin(β/) sin 1 (γ α), sin(β/) cos 1 (γ α), cos(β/) sin 1 (γ + α), () where ǫ = ±1 according to the following sign, ǫ sgn { cos 1 (γ + α) cos(β/)}, sin θ 0. (3) Remarkably, eq. () reduces to the correct results obtained above in the two subcases corresponding to θ = π, where cos(β/) = 0 and/or cos 1 (γ + α) = 0, respectively. Note that in the latter two subcases, ǫ as defined in eq. (3) is indeterminate. This is consistent One can can also determine ˆn up to an overall sign starting from eq. (18) by employing the relation R(ˆn, θ)ˆn = ˆn. The sign of ˆnsin θ can then be determined from eq. (7). 5

6 with the fact that the sign of ˆn is indeterminate when θ = π. Finally, one can easily verify that when θ = 0 [corresponding to cosβ = cos(γ + α) = 1], the direction of ˆn is indeterminate and hence arbitrary. One can rewrite the above results as follows. First, use eq. (0) to obtain: sin(θ/) = sin (β/) + cos (β/) cos 1 (γ + α), cos(θ/) = ǫ cos(β/) cos 1 (γ + α), (4) where we have used cos (θ/) = 1(1+cosθ) and sin (θ/) = 1 (1 cosθ). Since 0 θ π, it follows that 0 sin(θ/), cos(θ/) 1. Hence, the factor of ǫ defined by eq. (3) is required in eq. (4) to ensure that cos(θ/) is non-negative. In the mathematics literature, it is common to define the following vector consisting of four-components, q = (q 0, q 1, q, q 3 ), called a quaternion, as follows: ( ) q = cos(θ/), ˆn sin(θ/), (5) where the components of ˆn sin(θ/) comprise the last three components of the quaternion q and q 0 = ǫ cos(β/) cos 1 (γ + α), q 1 = ǫ sin(β/) sin 1 (γ α), q = ǫ sin(β/) cos 1 (γ α), q 3 = ǫ cos(β/) sin 1 (γ + α). (6) In the convention of 0 θ π, we have q Quaternions are especially valuable for representing rotations in computer graphics software. If one expresses ˆn in terms of a polar angle θ n and azimuthal angle φ n as in eq. (11), then one can also write down expressions for θ n and φ n in terms of the Euler angles α, β and γ. Comparing eqs. (11) and (), it follows that: tanθ n = (n 1 + n )1/ n 3 = ǫ tan(β/) sin 1 (γ + α) (7) where we have noted that (n 1 + n )1/ = sin(β/) 0, since 0 β π, and the sign ǫ = ±1 is defined by eq. (3). Similarly, cosφ n = n 1 (n 1 + n ) = ǫ sin 1 (γ α) = ǫ cos 1 (π γ + α), (8) or equivalently sin φ n = n (n 1 + n ) = ǫ cos 1 (γ α) = ǫ sin 1 (π γ + α), (9) φ n = 1 (ǫπ γ + α) mod π (30) 3 In comparing with other treatments in the mathematics literature, one should be careful to note that the convention of q 0 0 is not universally adopted. Often, the quaternion q in eq. (5) will be re-defined as ǫq in order to remove the factors of ǫ from eq. (6), in which case ǫq

7 Indeed, given that 0 α, γ < π and 0 β π, we see that θ n is determined mod π and φ n is determine mod π as expected for a polar and azimuthal angle, respectively. One can also solve for the Euler angles in terms of θ, θ n and φ n. First, we rewrite eq. (0) as: cos (θ/) = cos (β/) cos 1 (γ + α). (31) Then, using eqs. (7) and (31), it follows that: Plugging this result back into eqs. (7) and (31) yields sin(β/) = sin θ n sin(θ/). (3) ǫ sin 1 cosθ (γ + α) = ǫ cos 1 cos(θ/) (γ + α) = n sin(θ/) 1 sin θ n sin (θ/), (33) 1 sin θ n sin (θ/). (34) Note that if β = π then eq. (3) yields θ = π and θ n = π/, in which case γ + α is indeterminate. This is consistent with the observation that ǫ is indeterminate if cos(β/) = 0 [cf. eq. (3)]. We shall also make use of eqs. (8) and (9), ǫ sin 1 (γ α) = cosφ n, (35) ǫ cos 1 (γ α) = sin φ n, (36) Finally, we employ eqs. (34) and (35) to obtain (assuming β π): cos(θ/) sin φ n 1 sin θ n sin (θ/) = ǫ [ cos 1 (γ α) cos 1 (γ + α)] = ǫ sin(γ/) sin(α/). Since 0 1γ, 1 α < π, it follows that sin(γ/) sin(α/) 0. Thus, we may conclude that if γ 0, α 0 and β π then { } cos(θ/) ǫ = sgn sin φ n, (37) 1 sin θ n sin (θ/) If either γ = 0 or α = 0, then the argument of sgn in eq. (37) will vanish. In this case, sin 1 (γ + α) 0, and we may use eq. (33) to conclude that ǫ = sgn {cosθ n}, if θ n π/. The case of θ n = φ n = π/ must be separately considered and corresponds simply to β = θ and α = γ = 0, which yields ǫ = 1. The sign of ǫ is indeterminate if sin θ = 0 as noted below eq. (3). 4 The latter includes the case of β = π, which implies that θ = π and θ n = π/, where γ + α is indeterminate [cf. eq. (34)]. 4 In particular, if θ = 0 then θ n and φ n are not well-defined, whereas if θ = π then the signs of cosθ n, sin φ n and cosφ n are not well-defined [cf. eqs. (8) and (11)]. 7

8 There is an alternative strategy for determining the Euler angles in terms of θ, θ n and φ n. Simply set the two matrix forms for R(ˆn, θ), eqs. (1) and (18), equal to each other, where ˆn is given by eq. (11). For example, R 33 = cosβ = cosθ + cos θ n (1 cosθ). (38) where the matrix elements of R(ˆn, θ) are denoted by R ij. It follows that sin β = sin(θ/) sin θ n 1 sin θ n sin (θ/), (39) which also can be derived from eq. (3). Next, we note that if sin β 0, then sin α = R 3 sin β, cosα = R 13 sin β, sin γ = R 3 sin β, cosγ = R 31 sin β. Using eq. (1) yields (for sin β 0): sin α = cosθ n sin φ n sin(θ/) cosφ n cos(θ/) 1 sin θ n sin (θ/) cosα = cosθ n cos φ n sin(θ/) + sin φ n cos(θ/) 1 sin θ n sin (θ/) sin γ = cosθ n sin φ n sin(θ/) + cosφ n cos(θ/) 1 sin θ n sin (θ/) cosγ = cos θ n cosφ n sin(θ/) + sin φ n cos(θ/) 1 sin θ n sin (θ/), (40), (41), (4). (43) The cases for which sin β = 0 must be considered separately. Since 0 β π, sin β = 0 implies that β = 0 or β = π. If β = 0 then eq. (38) yields either (i) θ = 0, in which case R(ˆn, θ) = I and cos β = cos(γ + α) = 1, or (ii) sin θ n = 0, in which case cosβ = 1 and γ + α = θ mod π, with γ α indeterminate. If β = π then eq. (38) yields θ n = π/ and θ = π, in which case cosβ = 1 and γ α = π φ mod π, with γ + α indeterminate. One can use eqs. (40) (43) to rederive eqs. (33) (36). For example, if γ 0, α 0 and sin β 0, then we can employ a number of trigonometric identities to derive 5 cos 1 (γ ± α) = cos(γ/) cos(α/) sin(γ/) sin(α/) = sin(γ/) cos(γ/) sin(α/) cos(α/) sin (γ/) sin (α/) sin(γ/) sin(α/) = sin γ sin α (1 cosγ)(1 cosα) (1 cosγ) 1/ (1 cos α) 1/. (44) 5 Since sin(α/) and sin(γ/) are positive, one can set sin(α/) = { 1 [1 cos(α/)]} 1/ and sin(γ/) = { 1 [1 cos(γ/)]} 1/ by taking the positive square root in both cases, without ambiguity. 8

9 and sin 1 (γ ± α) = sin(γ/) cos(α/) ± cos(γ/) sin(α/) = = sin(γ/) sin(α/) cos(α/) sin(α/) sin(γ/) sinα sin(α/) ± sin γ sin(α/) ± sin(γ/) sin(γ/) cos(γ/) sin(α/) sin(γ/) 1 cos γ 1 cosα = 1 sin α 1 cosα ± 1 sin γ 1 cos γ = sin α(1 cosγ) ± sin γ(1 cos α) (1 cos α) 1/ (1 cosγ) 1/. (45) We now use eqs. (40) (43) to evaluate the above expressions. To evaluate the denominators of eqs. (44) and (45), we compute: (1 cos γ)(1 cosα) = 1 = sin φ n = ( sin φ n cos(θ/) 1 sin θ n sin (θ/) + sin φ n cos (θ/) cos θ n cos φ n sin (θ/) 1 sin θ n sin (θ/) sin φ n sin φ n cos(θ/) 1 sin θ n sin (θ/) + cos (θ/) 1 sin θ n sin (θ/) cos(θ/) 1 sin θ n sin (θ/)). Hence, (1 cosγ) 1/ (1 cosα) 1/ = ǫ ( sin φ n ) cos(θ/), 1 sin θ n sin (θ/) where ǫ = ±1 is the sign defined by eq. (37). Likewise we can employ eqs. (40) (43) to evaluate: [ ] cos(θ/) cos(θ/) sin γ sin α (1 cos γ)(1 cosα) = sin φ 1 sin θ n sin n, (θ/) 1 sin θ n sin (θ/) sin γ sin α + (1 cos γ)(1 cosα) = sin φ n [sin φ n sin α(1 cosγ) + sin γ(1 cosα) = ] cos(θ/), 1 sin θ n sin (θ/) [ cosθ n sin(θ/) sin φ 1 sin θ n sin n (θ/) sin α(1 cosγ) + sin γ(1 cosα) = cosφ n [sin φ n ] cos(θ/). 1 sin θ n sin (θ/) ] cos(θ/), 1 sin θ n sin (θ/) 9

10 Inserting the above results into eqs. (44) and (45), it immediately follows that cos 1 ǫ (γ + α) = sin 1 ǫ (γ + α) = cos(θ/) 1 sin θ n sin (θ/), cos 1 (γ α) = ǫ sin φ n, (46) cosθ n sin(θ/) 1 sin θ n sin (θ/), sin 1 (γ α) = ǫ cosφ n, (47) where ǫ is given by eq. (37). We have derived eqs. (46) and (47) assuming that α 0, γ 0 and sin β 0. Since cos(β/) is then strictly positive, eq. (3) implies that ǫ is equal to the sign of cos 1 (γ + α), which is consistent with the expression for cos 1 (γ + α) obtained above. Thus, we have confirmed the results of eqs. (33) (36). If α = 0 and/or γ = 0, then the derivation of eqs. (44) and (45) is not valid. Nevertheless, eqs. (46) and (47) are still true if sin β 0, as noted below eq. (37), with ǫ = sgn(cosθ n ) for θ n π/ and ǫ = +1 for θ n = φ n = π/. If β = 0, then as noted below eq. (43), either θ = 0 in which case ˆn is undefined, or θ 0 and sin θ n = 0 in which case the azimuthal angle φ n is undefined. Hence, β = 0 implies that γ α is indeterminate. Finally, as indicated below eq. (34), γ + α is indeterminate in the exceptional case of β = π (i.e., θ = π and θ n = π/). EXAMPLE: Suppose α = γ = 150 and β = 90. Then cos 1 (γ + α) = 1 3, which implies that ǫ = 1. Eqs. (0) and () then yield: cosθ = 1 4, ˆn = 1 5 (0,, 1). (48) The polar and azimuthal angles of ˆn [cf. eq. (11)] are then given by φ n = 90 (mod π) and tanθ n =. The latter can also be deduced from eqs. (7) and (30). Likewise, given eq. (48), one obtains cosβ = 0 (i.e. β = 90 ) from eq. (38), ǫ = 1 from eq. (37), γ = α from eqs. (35) and (36), and γ = α = 150 from eqs. (33) and (34). One can verify these results explicitly by inserting the values of the corresponding parameters into eqs. (1) and (18) and checking that the two matrix forms for R(ˆn, θ) coincide. References 1. W.-K. Tung, Group Theory in Physics (World Scientific, Singapore, 1985).. K.N. Srinivasa Rao, The Rotation and Lorentz Groups and their Representations for Physicists (Wiley Eastern Limited, New Delhi, 1988). 3. S.L. Altmann, Rotations, Quaternions, and Double Groups (Dover Publications, Inc., New York, NY, 005). 10